Question

A uniformly charged conducting sphere of \( 2.4\,m\) diameter has a surface charge density of \(80.0\,µCm^{-2}\) . 
(a) Find the charge on the sphere. 
(b) What is the total electric flux leaving the surface of the sphere?

Answer 1

(a) Diameter of the sphere, \(d = 2.4 m\)
Radius of the sphere,\(r = 1.2\,m\)
Surface charge density, \(σ = 80.0\,µCm^{-2} = 80 × 10^{−6} Cm^{-2}\)
Total charge on the surface of the sphere,
Q = Charge density × Surface area
\(= σ × 4πr^2\)
\(= 80 × 10^{−6} × 4 × 3.14 × (1.2)^2\)
\(= 1.447 × 10^{−3} C\)
Therefore, the charge on the sphere is \(1.447 × 10^{−3} C\).

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(b) Total electric flux ( \(\phi_ {Total}\)) leaving out the surface of a sphere containing net charge Q is given by the relation,
                                                \(\phi _{Total} =\frac{Q }{ ε_0}\)
Where,
\(ε_0\)= Permittivity of free space
\(= 8.854 × 10^{−12 }N^{−1} C^2 m^{−2}\)
\(Q = 1.447 × 10^{−3} C\)
                                              \(\phi_{Total} = \frac{1.447 × 10^{-3}}{ 8.854 × 10^ {-12}}\)
\(= 1.63 × 10^8 N C^{−1} m^2\)
Therefore, the total electric flux leaving the surface of the sphere is \(= 1.63 × 10^8 N C^{−1} m^2\).